# High current melting a spanner - what's happening?

Just a couple of guys doing fun things with a DIY low voltage very high current transformer. One of the things is putting a spanner on a brick and touching the two ends with an extremely thick copper cable carrying several thousand amps.

The spanner then becomes red hot and melts. And here we come to the question:

Why does the spanner turn red hot at the ends first and then later towards the center? I would have thought uniform current would have heated it evenly

• Before clicking the link "I bet it's photonic induction". [click] "Yep" – Connor Wolf Jun 18 '18 at 20:10
• Viewing the video I see that the narrowest parts of the wrench heat up first. This is entirely expected. – Hot Licks Jun 19 '18 at 0:53

There is heating from the contact points, but not enough to make them turn red. More heat comes from the thin section. Where both sources heat the metal it gets hotter than the rest of the thin section, causing the resistance to rise as it heats, yielding more localized heating (positive feedback), and so on, so the ends of the thin section get hot first and the hot area propagates toward the center of the thin section.

It may only take a relatively small temperature difference to start the positive feedback in a given section. See, for example, this curve.

• Wouldn't that be a negative feedback loop? As the temperature increases the resistivity increases as well. – Todd Sewell Jun 19 '18 at 13:07
• @ToddSewell Consider a cross-sectional slice area A of the wrench of length x. A current I is flowing. The power dissipated in that segment is I^2*rho*x/A. The higher the resistivity rho and the lower the cross-sectional area the more power is dissipated in that slice. Or to look at it another way if you put a 1 ohm resistor in series with a 1K resistor across the mains, the 1 ohm will stay cool and the 1K will get very hot. – Spehro Pefhany Jun 19 '18 at 13:21
• Ah of course, we need to look at power, I was thinking about current. Thanks! – Todd Sewell Jun 19 '18 at 13:24
• @ToddSewell because of $$P=RI^2$$ and the current is constant, the feedback loop is positive. – Crowley Jun 19 '18 at 14:40

The current density where you make contact is much greater than the current density a couple of cm further into the spanner/wrench. That's one point.

The contact resistance is much greater where the copper wires make contact.

Both these points make the wrench get hotter at the ends first.

The highest resistance is, initially, at the points where your conductors connect. As a gross general rule, high carbon steel has a slightly Negative Temperature Coefficient (NTC) of Resistance, meaning the resistance decreases as temperature increases, so once the wrench heats up, resistance drops across the entire length to a more uniform level.

The Ohm's law works there in one of its most educational ways.

Joule's heat can be calculated as $$P=UI$$ where U is the voltage drop over the part and I is a current through it.

Ohm's law says $$R=\frac UI.$$

Putting this together we know that high current power source was used. The resistance and current are therefore known and we have enough information to estimate the heating power as $$P=RI^2.$$

The highest resistance is at the contact between the spanner and the clamps and the crossection is lowest there as well, that's why the glowing started there and propagated through whole spanner.

That means:

• the higher current, the higher heating power and thus higher temperature
• the higher resistance, the higher heating power. (One need to provide higher voltage to sustain same current)